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Math Help - Quadratic eqn

  1. #1
    Member helloying's Avatar
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    Quadratic eqn

    The roots of the eqn x^2-8x+h
    are (alpha) and (alpha+3k)

    Express h in terms of k
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by helloying View Post
    The roots of the eqn x^2-8x+h
    are (alpha) and (alpha+3k)

    Express h in terms of k
    For
    Quadratic equation

    ax^2 +bx +c =0

    Sum of roots of quadratic equation = \frac{-b}{a}

    Product of roots of equation = \frac{c}{a}.....(2)


    Hence using (2) we get

    <br />
(\alpha) \times (\alpha + 3k) = h --------(3)

    And

    (\alpha) + (\alpha+3k) = 8

    \alpha= (8-3k)/2 --------------(1)

    Use one in (3)

    \frac{(8-3k)}{2} \times \frac{8+3k}{2} = h

    64- 9k^2 = 4h
    Last edited by ADARSH; March 18th 2009 at 03:36 AM. Reason: typo
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  3. #3
    MHF Contributor
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    Quote Originally Posted by helloying View Post
    The roots of the eqn x^2-8x+h
    are (alpha) and (alpha+3k)

    Express h in terms of k
    I will guess that the author of the assignment forgot the "equals zero" part of the "equation". (Without the "equals" sign, it's not an "equation"; without the assumption that the other side was "zero", there's no way to find the roots.)

    In case you haven't learned about the formulas relating the roots of a quadratic function to the coefficients, you can instead use the relationship between roots and factors to find the factorization:

    . . . . . \mbox{If }\, x\, =\, a\, \mbox{ is a root, then }\, x\, -\, a\, \mbox{ is a factor.}

    In your case, the zeroes are:

    . . . . . x\, =\, \alpha\, \mbox{ and }\, x\, =\, \alpha\, +\, 3k

    Then the factors must have been:

    . . . . . x\, -\, \alpha\, \mbox{ and }\, x\, -\, (\alpha\, +\, 3k)

    Multiplying, we get:

    . . . . . (x\, -\, \alpha)(x\, -\, (\alpha\, +\, 3k))\, =\, x^2\, +\, (-2\alpha\, -\, 3k)x\, +\, (-\alpha^2\, -\, 3\alpha k)

    Comparing coefficients, we see that:

    . . . . . -2\alpha\, -\, 3k\, =\, -8

    Then:

    . . . . . \alpha\, =\, 4\, -\, \frac{3}{2}k

    Comparing coefficients again, we see that:

    . . . . . -\alpha^2\, -\, 3\alpha k\, =\, h

    Substituting from above, we get:

    . . . . . -\left(4\, -\, \frac{3}{2}k\right)^2\, -\, 3\left(4\, -\, \frac{3}{2}k\right)k\, =\, h

    . . . . . -16\, +\, 12k\, -\, \frac{9}{4}k^2\, -\, 12k\, +\, \frac{9}{2}k^2\, =\, h

    ...which simplifies to give a result equivalent to the previous reply.
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